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| author | kinitrupti | 2017-05-12 18:53:46 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:53:46 +0530 |
| commit | f270f72badd9c61d48f290c3396004802841b9df (patch) | |
| tree | bc8ba99d85644c62716ce397fe60177095b303db /Engineering_Physics_by_P._V._Naik/Chapter11.ipynb | |
| parent | 64d949698432e05f2a372d9edc859c5b9df1f438 (diff) | |
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diff --git a/Engineering_Physics_by_P._V._Naik/Chapter11.ipynb b/Engineering_Physics_by_P._V._Naik/Chapter11.ipynb new file mode 100755 index 00000000..39762859 --- /dev/null +++ b/Engineering_Physics_by_P._V._Naik/Chapter11.ipynb @@ -0,0 +1,620 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1d5f4970753c62d94a2bd202867cc0f79046f1baac4b1a42721a5ae6844ad5f4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "11: Nuclear Radiations and Detectors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.1, Page number 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "r0=1.2; #radius(fm)\n",
+ "A=7; #mass number \n",
+ "\n",
+ "#Calculation \n",
+ "r=r0*A**(1/3);\t #radius of Li(fm) \n",
+ "\n",
+ "#Result\n",
+ "print \"The radius of Li is\",round(r,4),\"fm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius of Li is 2.2955 fm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.2, Page number 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "M=235.043945; #atomic mass of uranium(u)\n",
+ "Z=92; #atomic number of uranium\n",
+ "mp=1.007825; #mass of proton(kg)\n",
+ "N=143; #no.of neutrons\n",
+ "mn=1.008665; #mass of neutron(kg)\n",
+ "A=235; #number of nucleons\n",
+ "\n",
+ "#Calculation \n",
+ "B=(((Z*mp)+(N*mn)-(M))/A)*931.5; #Binding energy(MeV)\n",
+ "\n",
+ "#Result\n",
+ "print \"The binding energy per nucleon is\",round(B,3),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The binding energy per nucleon is 7.591 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.3, Page number 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "#After removing one neutron from Ca(A=43;Z=20) it becomes Ca(A=42;Z=20)\n",
+ "M=41.958622; #mass of Ca(A=42;Z=20)(kg)\n",
+ "mn=1.008665; #mass of neutron(kg)\n",
+ "E=42.95877; #mass of Ca(A=43;Z=20)(kg)\n",
+ "\n",
+ "#Calculation \n",
+ "C=M+mn;\n",
+ "D=C-E;\n",
+ "B=D*931.5; #Binding energy of neutron(MeV)\n",
+ "\n",
+ "#Result\n",
+ "print \"The binding energy of neutron is\",round(B,4),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The binding energy of neutron is 7.9336 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.4, Page number 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "mBe=9.012182; #Atomic mass of beryllium(u)\n",
+ "mHe=4.002603; #Atomic mass of helium\n",
+ "mn=1.008665; #mass of neutron(kg)\n",
+ "mC=12.000000; #Atomic mass of carbon\n",
+ "\n",
+ "#Calculation \n",
+ "Q=(mBe+mHe-mn-mC)*931.5 #energy balance of the reaction(MeV)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Q-value is\",round(Q,1),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Q-value is 5.7 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.5, Page number 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "mLi=7.016004; #mass of Lithium(A=7)(u)\n",
+ "mH=1.007825; #mass of Hydrogen(A=1)(u)\n",
+ "mHe=4.002603; #mass of helium(A=4)(u)\n",
+ "p=0.5; #energy of proton(MeV)\n",
+ "\n",
+ "#Calculation \n",
+ "Q=(mLi+mH-2*(mHe))*931.5 #energy balance of the reaction(MeV)\n",
+ "#The energy of 2 alpha particles is equal to the Q-value + energy of proton.\n",
+ "Ealpha=(Q+p)/2; #energy of each alpha particle(MeV)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Q-value of the reaction is\",round(Q,2),\"MeV\"\n",
+ "print \"The energy of each alpha particle is\",round(Ealpha,3),\"MeV\"\n",
+ "print \"answer for energy in the book varies due to rounding off errors\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Q-value of the reaction is 17.35 MeV\n",
+ "The energy of each alpha particle is 8.924 MeV\n",
+ "answer for energy in the book varies due to rounding off errors\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.6, Page number 228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "wt=1000; #weight(gm)\n",
+ "A=235; #mass number of uranium\n",
+ "N=(6.02*10**23/A)*wt; #no.of nuclei in 1kg of uranium\n",
+ "Q=208; #energy-balance of the reaction\n",
+ "\n",
+ "#Calculation \n",
+ "E=N*Q; #Energy released(MeV)\n",
+ "#1MeV=4.45*10^-20kWh\n",
+ "E=E*4.45*10**-20;\n",
+ "\n",
+ "#Result\n",
+ "print \"The energy released is\",round(E/10**7,3),\"*10**7 kWh\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy released is 2.371 *10**7 kWh\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.7, Page number 228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "wt=5000; #weight(gm)\n",
+ "A=235; #mass number of uranium\n",
+ "Ef=208; #Energy released per fission(MeV)\n",
+ "\n",
+ "#Calculation \n",
+ "N=(6.02*10**23/A)*wt; #number of nuclei in 5 Kg\n",
+ "E=N*Ef; #Energy(MeV)\n",
+ "E=E*1.6*10**-13; #Energy(J)\n",
+ "T=24*60*60; #time\n",
+ "P=E/T; #power(MW)\n",
+ "\n",
+ "#Result\n",
+ "print \"The power output of a nuclear reactor is\",round(P/10**6),\"MW\"\n",
+ "print \"answer given in the book is wrong\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The power output of a nuclear reactor is 4934.0 MW\n",
+ "answer given in the book is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.8, Page number 228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "A=235; #mass number of uranium\n",
+ "p=1000; #amount of electric power produced(MW)\n",
+ "e=0.32; #energy conversion efficiency of the plant\n",
+ "f=200; #fission energy per event(MeV)\n",
+ "\n",
+ "#Calculation \n",
+ "I=p/e; #Input nuclear energy(MW)\n",
+ "TE=I*(10**6)*3600*24*365; #total energy(J)\n",
+ "EF=f*(10**6)*1.6*10**-19; #Energy released per fission event(J)\n",
+ "N=TE/EF; #Number of nuclei required\n",
+ "M=N*A/(6.02*10**23); #corresponding mass(g)\n",
+ "\n",
+ "#Result\n",
+ "print \"The amount of uranium required is\",round(M*10**-3,1),\"kg\"\n",
+ "print \"answer in the book varies due to rounding off errors\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The amount of uranium required is 1202.2 kg\n",
+ "answer in the book varies due to rounding off errors\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.9, Page number 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "q=1.6*10**-19; #charge of the particle(c)\n",
+ "B=1; #magnetic field(T)\n",
+ "m=1.67*10**-27; #mass of proton(kg)\n",
+ "r=0.5; #radius(m)\n",
+ "\n",
+ "#Calculation \n",
+ "omega=(q*B)/m; #angular frequency(radian/s)\n",
+ "v=(omega/(2*math.pi))*10**-8; #frequency(MHz)\n",
+ "s=omega*r; #speed of proton(m/s)\n",
+ "K=(m*(s**2))*(1/2)*6.27*10**12; #kinetic energy of protons emerging from cyclotron(MeV)\n",
+ "\n",
+ "#Result\n",
+ "print \"The frequency of oscillator to accelerate protons is\",round(omega/10**8,2),\"*10**8 radian/s\"\n",
+ "print \"The speed of proton is\",round(s/10**7,1),\"*10**7 m/s\"\n",
+ "print \"The kinetic energy of protons emerging from the cyclotron is\",int(K),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequency of oscillator to accelerate protons is 0.96 *10**8 radian/s\n",
+ "The speed of proton is 4.8 *10**7 m/s\n",
+ "The kinetic energy of protons emerging from the cyclotron is 12 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.10, Page number 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=1.83*10**17; #average density of carbon nucleus(kg/m^3)\n",
+ "m=12; #mass(u)\n",
+ "e=1.66*10**-27;\n",
+ "\n",
+ "#Calculation \n",
+ "r=(m*e/((4/3)*math.pi*rho))**(1/3)*10**15; #radius(fm)\n",
+ "\n",
+ "#Result\n",
+ "print \"The radius is\",round(r,2),\"fm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius is 2.96 fm\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.11, Page number 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "q=1.6*10**-19; #charge of the particle(c)\n",
+ "B=5; #magnetic field(T)\n",
+ "m=9.1*10**-31; #mass of electron(kg)\n",
+ "\n",
+ "#Calculation \n",
+ "v=(q*B)/(2*math.pi*m); #cyclotron frequency(Hz)\n",
+ "\n",
+ "#Result\n",
+ "print \"cyclotron frequency of electron is\",round(v/10**11,1),\"*10**11 Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "cyclotron frequency of electron is 1.4 *10**11 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.12, Page number 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "k=1.5; #maximum kinetic energy(MeV)\n",
+ "m=1.67*10**-27; #mass of proton(kg)\n",
+ "q=1.6*10**-19; #charge of particle(c)\n",
+ "r=0.35; #radius(m)\n",
+ "\n",
+ "#Calculation \n",
+ "B=math.sqrt(k*10**6*q*2*m)/(q*r); #magnetic field(T)\n",
+ "\n",
+ "#Result\n",
+ "print \"The mgnetic field is\",round(B,1),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mgnetic field is 0.5 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.13, Page number 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "m=1.67*10**-27; #mass of proton(kg)\n",
+ "q=1.6*10**-19; #charge of particle(q)\n",
+ "v=25; #cyclotron frequency(MHz)\n",
+ "\n",
+ "#Calculation \n",
+ "B=(v*10**6*2*math.pi*m)/q; #magnetic field(T)\n",
+ "\n",
+ "#Result\n",
+ "print \"The required magnetic field is\",round(B,4),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required magnetic field is 1.6395 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example number 11.14, Page number 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "v=20; #cyclotron frequency(MHz)\n",
+ "B=1.3; #magnetic field(T)\n",
+ "\n",
+ "#Calculation \n",
+ "d=2*math.pi*v*10**6/B; #charge to mass ratio of proton(C/kg)\n",
+ "\n",
+ "#Result\n",
+ "print \"charge to mass ratio of proton is\",round(d/10**6,2),\"*10**6 C/kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "charge to mass ratio of proton is 96.66 *10**6 C/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |